$\underset{n}{lim}{b}_n=\underset{n-2n}{lim}\frac{2n}{3n-4}=\underset{n}{lim}\frac{2}{2}=$

$$

so condition $($ii$)$ in the Alternating Series Test, that the limit of the terms approaches $a_{,}$ is not satisfied. Instead, we lack at the limit of the nth term of the series:

$\underset{n}{lim}{a}_n=\underset{n}{lim}\frac{(-1{)}^{n}2n}{3n-4}$

This limit does not exist, so the series diverges by the Test for Divergence.